JEE PYQ: Mathematics In Physics Question 2
Question 2 - 2021 (18 Mar Shift 1)
The time period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$. The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of ‘$g$’ using this pendulum is ‘$x$’. The value of ‘$x$’ to the nearest integer is:-
(1) 2%
(2) 3%
(3) 5%
(4) 4%
Show Answer
Answer: (2)
Solution
$g = \frac{4\pi^2 \ell}{T^2}$ $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta T}{T} = \frac{0.1}{10} + 2\left(\frac{1}{200 \times 0.5}\right)$ $\frac{\Delta g}{g} = \frac{1}{100} + \frac{1}{50}$ $\frac{\Delta g}{g} \times 100 = 3%$
BETA
